Em nhờ tí

Ta có $\lim_{x\rightarrow 0}\dfrac{\sin x}{x}=1$
và $f\left( 0 \right)=0$
Xét
$\lim_{x\rightarrow 0^{-}}f\left(x\right)
\lim_{x\rightarrow 0^{-}} f\left(x\right)=
\lim_{x\rightarrow 0^{-}}\dfrac{\sqrt{1+2x}-1+\sin ^{2}x}{x}
=\lim_{x\rightarrow 0^{-}}\dfrac{\left(\sqrt{1+2x}-1\right).\left(\sqrt{1+2x}+1\right)}{x.\left(\sqrt{1+2x}+1\right)}+\lim_{x\rightarrow 0^{-}}\dfrac{\sin ^{2}x}{x}
=\lim_{x\rightarrow 0^{-}}\dfrac{2}{\left(\sqrt{1+2x}+1\right)}+\lim_{x\rightarrow 0^{-}}\dfrac{\sin x.\sin x}{x}=1+0=1$
 

Chuyên mục

$\lim_{x\rightarrow 0^{-}}f\left(x\right)
=\lim_{x\rightarrow 0^{-}}\dfrac{\sqrt{1+2x}-1+\sin ^{2}x}{x}$
$=\lim_{x\rightarrow 0^{-}}\dfrac{\left(\sqrt{1+2x}-1\right).\left(\sqrt{1+2x}+1\right)}{x.\left(\sqrt{1+2x}+1\right)}+\lim_{x\rightarrow 0^{-}}\dfrac{\sin ^{2}x}{x}$
$=\lim_{x\rightarrow 0^{-}}\dfrac{2}{\left(\sqrt{1+2x}+1\right)}+\lim_{x\rightarrow 0^{-}}\dfrac{\sin x.\sin x}{x}=1+0=1$
 

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